Week 2 Worksheet - Solutions

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Functions

1. Write $z=x+iy$. For $f(z)=\mathsf{Re}(z)$ we have $u(x,y)=x$ and $v(x,y)=0$. For $$f(z)=|z|=\sqrt{x^2+y^2}$$ we have $u(x,y)=\sqrt{x^2+y^2}$ and $v(x,y)=0$.

2. 1. $(x+iy{)}^2=x^2-y^2+2xyi$ so $u(x,y)=x^2-y^2$ and $v(x,y)=2xy$.
   2. $(\mathrm{\nabla }u)(x,y)=⟨2x,-2y⟩$ and $(\mathrm{\nabla }v)(x,y)=⟨2y,2x⟩$.

   3. Calculate $$\begin{array}{rl}& (\mathrm{\nabla }u)(x,y)\bullet (\mathrm{\nabla }v)(x,y)\\ =& ⟨2x,-2y⟩\bullet ⟨2y,2x⟩\\ =& (2x)(2y)+(-2y)(2x)\\ =& 0\end{array}$$

3. 1. $u(x,y)={\displaystyle \frac{x}{x^2+y^2}}$ and $v(x,y)={\displaystyle \frac{-y}{x^2+y^2}}$
   2. Fix $k\in \mathbb{R}$ non-zero. The equation $u(x,y)=k$ has the form $$\begin{array}{rl}& {\displaystyle \frac{x}{x^2+y^2}}=k\\ \Rightarrow & k{x}^2-x+y^2=0\\ \Rightarrow & k{(x-\frac{1}{2k})}^2-\frac{1}{4k}+k{y}^2=0\end{array}$$ so we have $${(x-\frac{1}{2k})}^2+y^2={\left(\frac{1}{2k}\right)}^2$$ which is a circle centered at $(\frac{1}{2k},0)$ of radius $\frac{1}{2k}$.
4. Fix $a\in \mathbb{C}$ non-zero. We need to prove that $$\underset{z\to a}{lim}{\displaystyle \frac{1}{z}}={\displaystyle \frac{1}{a}}$$ holds. Fix $ϵ>0$. If $|z-a|<|a|/2$ then $||z|-|a||<|a|/2$ so $|z|>|a|/2$. We then have $$|\frac{1}{z}-\frac{1}{a}|=\left|{\displaystyle \frac{z-a}{za}}\right|\le {\displaystyle \frac{2}{|a{|}^2}}|z-a|$$ which will be at most $ϵ$ if $|z-a|=ϵ|a{|}^2/4$. Take $\delta =min\{|a|/2,ϵ|a{|}^2/4\}$.

5. 1. With $z=t$ we have $$\underset{t\to 0}{lim}{\displaystyle \frac{|t|}{t}}=1$$ and with $z=it$ we have $$\underset{t\to 0}{lim}{\displaystyle \frac{|it|}{it}}=-i$$ so the complex limit does not exist.

   2. Here $\left|{\displaystyle \frac{|z{|}^2}{z}}\right|=|z|$ which does go to zero as $z\to 0$.

   3. With $z=t>0$ we have $$\underset{t\to 0}{lim}\mathsf{Arg}(t)=0$$ while with $z=it$ we have $$\underset{t\to 0}{lim}\mathsf{Arg}(t)={\displaystyle \frac{\pi }{2}}$$ so the complex limit does not exist.

6. 1. One can use the fact that $|\mathsf{Re}(z)-\mathsf{Re}(a)|\le |z-a|$ to prove $$\underset{z\to a}{lim}\mathsf{Re}(z)=\mathsf{Re}(a)$$ for all $a\in \mathbb{C}$. Therefore $f$ is continuous on $\mathbb{C}$.

   2. Use the reverse triangle inequality $||z|-|a||\le |z-a|$ to prove that $$\underset{z\to a}{lim}|z|=|a|$$ holds.

   3. Fix $a\in \mathbb{C}\setminus (-\mathrm{\infty },0]$ and put $\theta =\mathsf{Arg}(a)$.

Holomorphic Functions

7. 1. We calculate $f(z+h)=z^2+2zh+h^2+z+h$ so that $${\displaystyle \frac{f(z+h)-f(z)}{h}}={\displaystyle \frac{2zh+h^2+h}{h}}=2z+h+1$$ and $$\underset{h\to 0}{lim}{\displaystyle \frac{f(z+h)-f(z)}{h}}=\underset{h\to 0}{lim}2z+h+1=2z+1$$ for all $z\in \mathbb{C}$.

   2. We calculate $$\begin{array}{rl}{\displaystyle \frac{f(z+h)-f(z)}{h}}& ={\displaystyle \frac{{\displaystyle \frac{1}{z+h}}-{\displaystyle \frac{1}{z}}}{h}}={\displaystyle \frac{{\displaystyle \frac{-h}{(z+h)z}}}{h}}={\displaystyle \frac{-1}{(z+h)z}}\end{array}$$ and $$\underset{h\to 0}{lim}{\displaystyle \frac{f(z+h)-f(z)}{h}}=\underset{h\to 0}{lim}{\displaystyle \frac{-1}{(z+h)z}}=-{\displaystyle \frac{1}{z^2}}$$ for all $z\in \mathbb{C}\setminus \{0\}$.

8. 1. From the definition $$\underset{h\to 0}{lim}{\displaystyle \frac{f(h)-f(0)}{h}}={\displaystyle \frac{|h{|}^2}{h}}=\underset{h\to 0}{lim}{\displaystyle \frac{h\bar{h}}{h}}=\underset{h\to 0}{lim}\bar{h}=0$$ so the function has a derivative of zero at the origin.

   2. Fix $z\in \mathbb{C}$. The limit $$\begin{array}{rl}\underset{h\to 0}{lim}{\displaystyle \frac{f(z+h)-f(z)}{h}}& =\underset{h\to 0}{lim}{\displaystyle \frac{(z+h)\overline{(z+h)}-z\bar{z}}{h}}\\ & =\underset{h\to 0}{lim}{\displaystyle \frac{z\bar{h}+\bar{z}h+|h{|}^2}{h}}\end{array}$$ does not exist because if $h=t$ is real we get $$\underset{t\to 0}{lim}{\displaystyle \frac{zt+\bar{z}t+t^2}{t}}=z+\bar{z}$$ whereas if $h=it$ is imaginary we get $$\underset{t\to 0}{lim}{\displaystyle \frac{zit-\bar{z}it+t^2}{it}}=\bar{z}-z$$ so $f$ is not differentiable at any non-zero $z$. It is therefore not holomorphic on any domain.

Cauchy-Riemann Equations

9. From the definition $$\begin{array}{rl}({\partial }_{1}h)(x,y)& =\underset{t\to 0}{lim}{\displaystyle \frac{h(x+t,y)-h(x,y)}{t}}\\ & =\underset{t\to 0}{lim}{\displaystyle \frac{2(x+t)y-2xy}{t}}=\underset{t\to 0}{lim}2y=2y\end{array}$$ and $$\begin{array}{rl}({\partial }_{2}h)(x,y)& =\underset{t\to 0}{lim}{\displaystyle \frac{h(x,y+t)-h(x,y)}{t}}\\ & =\underset{t\to 0}{lim}{\displaystyle \frac{2x(y+t)-2xy}{t}}=\underset{t\to 0}{lim}2x=2x\end{array}$$ as expected from rules of differentiation.

10. From $$\frac{1}{z}=\frac{x-iy}{x^2+y^2}=\frac{x}{x^2+y^2}+i\frac{(-y)}{x^2+y^2}$$ we get $u(x,y)$ and $v(x,y)$. The partial derivatives are $$\begin{array}{rl}({\partial }_{1}u)(x,y)& ={\displaystyle \frac{(x^2+y^2)-2x(x)}{(x^2+y^2{)}^2}}={\displaystyle \frac{y^2-x^2}{(x^2+y^2{)}^2}}\\ ({\partial }_{2}u)(x,y)& ={\displaystyle \frac{-2xy}{(x^2+y^2{)}^2}}\end{array}$$ and $$\begin{array}{rl}({\partial }_{1}v)(x,y)& ={\displaystyle \frac{2xy}{(x^2+y^2{)}^2}}\\ ({\partial }_{2}u)(x,y)& ={\displaystyle \frac{-(x^2+y^2)+2y(y)}{(x^2+y^2{)}^2}}={\displaystyle \frac{y^2-x^2}{(x^2+y^2{)}^2}}\end{array}$$ so the Cauchy-Riemann equations are satisfied.

11. For this function $u(x,y)=\sqrt{x^2+y^2}$ and $v(x,y)=0$. Thus $$({\partial }_{1}u)(x,y)={\displaystyle \frac{x}{\sqrt{x^2+y^2}}}({\partial }_{2}u)(x,y)={\displaystyle \frac{y}{\sqrt{x^2+y^2}}}$$ and $({\partial }_{1}v)(x,y)=({\partial }_{2}v)(x,y)=0$ for all $(x,y)\ne (0,0)$. The Cauchy- Riemann equations are therefore not satisfied at any non-zero complex number.

    Since the partial derivatives of $u$ do not even exist at $(0,0)$ the function $f$ is certainly not differentiable there.

12. We calculate $$\begin{array}{rl}({\partial }_{1}u)(x,y)& =3x^2-3y^2\\ ({\partial }_{2}u)(x,y)& =-6xy\\ ({\partial }_{1}v)(x,y)& =6xy\\ ({\partial }_{2}v)(x,y)& =3x^2-3y^2\end{array}$$ and see that the equations are satisfied for all $(x,y)$.

    From $$u(x,y)+iv(x,y)=x^3+3i{x}^{2}y-3x{y}^2-y^3$$ we see that $$f(z)=z^3=(x+iy{)}^3$$ has the above real and imaginary parts.

13. We write $${\displaystyle \frac{1}{(x+iy{)}^4}}={\displaystyle \frac{(x-iy{)}^4}{(x^2+y^2{)}^4}}={\displaystyle \frac{x^4-4i{x}^{3}y-6x^2{y}^2+4ix{y}^3+y^4}{(x^2+y^2{)}^4}}$$ so that $u$ and $v$ are the real and imaginary parts of $f(z)=\frac{1}{z^4}$ on $\mathbb{C}\setminus \{0\}$. Since $f$ is holomorphic on that domain, the Cauchy-Riemann equations are satisfied for all $z$ therein.

14. We calculate $$\begin{array}{rl}{\partial }_1({\partial }_{1}u)+{\partial }_2({\partial }_{2}u)& ={\partial }_1({\partial }_{2}v)+{\partial }_2(-{\partial }_{1}v)\\ & ={\partial }_1({\partial }_{2}v)-{\partial }_2({\partial }_{1}v)=0\end{array}$$ by Clairaut's theorem.

15. From an earlier problem we have $$u(x,y)=x^3-3x{y}^{2}v(x,y)=3x^{2}y-y^3$$ and can calculate that $$\begin{array}{rl}{\partial }_1({\partial }_{1}u)+{\partial }_2({\partial }_{2}u)& ={\partial }_1(3x^2-3y^2)+{\partial }_2(-6xy)=6x-6x=0\\ {\partial }_1({\partial }_{1}v)+{\partial }_2({\partial }_{2}v)& ={\partial }_1(3x^2-3y^2)+{\partial }_2(-6xy)=6x-6x=0\end{array}$$ as desired.

16. We have $$\begin{array}{rl}({\partial }_{1}u)(x,y)& =5x^4-30x^2{y}^2+5y^4\\ ({\partial }_{2}u)(x,y)& =-20x^{3}y+20x{y}^3\end{array}$$ so $$\begin{array}{rl}({\partial }_{1}v)(x,y)& =20x^{3}y-20x{y}^3\\ ({\partial }_{2}v)(x,y)& =5x^4-30x^2{y}^2+5y^4\end{array}$$ from the Cauchy-Riemann equations. Partially integrating gives $$5x^{4}y-10x^2{y}^3+y^5+h(x)=v(x,y)=5x^{4}y-10x^2{y}^3+g(y)$$ which will be satisfied if e.g. $h(x)=0$ and $g(y)=y^5$. Then $$f(x+iy)=(x^5-10x^3{y}^2+5x{y}^4)+i(5x^{4}y-10x^2{y}^3+y^5)=(x+iy{)}^5$$ and $f(z)=z^5$.

17. We have $$\begin{array}{rl}({\partial }_{1}u)(x,y)& =3x^2-k{y}^2+12y-12\\ ({\partial }_{2}u)(x,y)& =-2kxy+12x\end{array}$$ so $$\begin{array}{rl}({\partial }_{1}v)(x,y)& =2kxy-12x\\ ({\partial }_{2}v)(x,y)& =3x^2-k{y}^2+12y-12\end{array}$$ from the Cauchy-Riemann equations. Partially integrating gives $$3x^{2}y-\frac{k}{3}{y}^3+6y^2-12y+h(x)=v(x,y)=k{x}^{2}y-6x^2+g(y)$$ and therefore $$3x^{2}y=k{x}^{2}y$$ if $h(x)=-6x^2$ and $g(y)=-\frac{k}{3}{y}^3+6y^2-12y$. Therefore, the only value of $k$ for which $u$ could possibly be the real part of a holomorphic function is $3$. When $k=3$ we have $$v(x,y)=3x^{2}y-6x^2-y^3+6y^2-12y$$ and $u,v$ are the real and imaginary parts of $f(z)=z^3+6i{z}^2-12z$.

18. If $u$ is constant and the real part of a holomorphic function then the Cauchy-Riemann equations would force $({\partial }_{1}v)(x,y)=0=({\partial }_{2}v)(x,y)$ whence $v$ is constant as well.

19. In the purported situation the Cauchy-Riemann equations become $$\begin{array}{c}{u}^{\prime }(x)=({\partial }_{1}u)(x,y)=({\partial }_{2}v)(x,y)=v^{\prime }(y)\\ 0=({\partial }_{2}u)(x,y)=-({\partial }_{1}v)(x,y)=0\end{array}$$ which tells us that $u^{\prime }$ and $v^{\prime }$ are constant. But then $u(x)=\mu x+b$ and $v(y)=\nu y+c$. Put $\lambda =\mu +i\nu$ and $a=b+ic$.

20. If we apply ${\partial }_1$ to the equation we get $$0=2({\partial }_{1}u)(x,y)+({\partial }_{1}v)(x,y)=2({\partial }_{2}v)(x,y)+({\partial }_{1}v)(x,y)$$ from the Cauchy-Riemann equations. We get $$0=2({\partial }_{2}u)(x,y)+({\partial }_{2}v)(x,y)=-2({\partial }_{1}v)(x,y)+({\partial }_{2}v)(x,y)$$ if we apply instead ${\partial }_2$. Adding twice the first equation to the second gives $5({\partial }_{2}v)(x,y)=0$ for all $(x,y)$. It then follows from the second equation that $({\partial }_{1}v)(x,y)=0$ for all $(x,y)$. The Cauchy-Riemann equations then gives $({\partial }_{1}u)(x,y)=0$ and $({\partial }_{2}u)(x,y)=0$ for all $(x,y)$. We conclude that $u$ and $v$ are constant, so that $f$ is as well.